Lowbit Sum 规律

Lowbit Sum

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

SubmitStatus

Problem Description

long long ans = 0;
for(int i = 1; i <= n; i ++)
    ans += lowbit(i)
lowbit(i)的意思是将i转化成二进制数之后，只保留最低位的1及其后面的0，截断前面的内容，然后再转成10进制数
比如lowbit(7),7的二进制位是111，lowbit(7) = 1
6 = 110(2),lowbit(6) = 2,同理lowbit(4) = 4,lowbit(12) = 4,lowbit(2) = 2,lowbit(8) = 8

每输入一个n，求ans

Input

多组数据，每组数据一个n(1 <= n <= 10^9)

Output

每组数据输出一行，对应的ans

Sample Input

1
2
3

Sample Output

1
3
4

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 #include <math.h>
 #include <algorithm>
 using namespace std;
 #define ll long long
 ll solve(int x)
 {
     if(x==)return ;
     return ((x+)>>)+(solve(x/)<<);
 }
 int main()
 {
     int n;
     while(~scanf("%d",&n))
     {
         printf("%lld\n",solve(n));
     }
 }